Problem 22 - Single Variable Calculus

Question

Find the derivative of the function. \(f(x)=5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\)

Step-by-Step Solution

Answer

The derivative of the function \(f(x) = 5x^{\frac{4}{3}} - \frac{2}{3}x^{\frac{3}{2}} + x^2 - 3x + 1\) is: \(f'(x) = \frac{20}{3}x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2x - 3\)

Step 1: Differentiate the first term of the function

For the first term, \(5x^{\frac{4}{3}}\), apply the power rule: $$\frac{d}{dx} \left(5x^{\frac{4}{3}}\right) = 5\left(\frac{4}{3}\right)x^{\frac{4}{3} - 1}= \frac{20}{3}x^{\frac{1}{3}}$$

Step 2: Differentiate the second term of the function

For the second term, \(-\frac{2}{3}x^{\frac{3}{2}}\), apply the power rule: $$\frac{d}{dx} \left(-\frac{2}{3}x^{\frac{3}{2}}\right) = -\frac{2}{3}\left(\frac{3}{2}\right)x^{\frac{3}{2} - 1} = -x^{\frac{1}{2}}$$

Step 3: Differentiate the third term of the function

For the third term, \(x^2\), apply the power rule: $$\frac{d}{dx}(x^2) = 2x^{2-1} = 2x$$

Step 4: Differentiate the fourth term of the function

For the fourth term, \(-3x\), apply the power rule: $$\frac{d}{dx}(-3x) = -3x^{1 - 1}= -3x^0 = -3$$

Step 5: Differentiate the fifth term of the function

For the fifth term, \(1\), as it is a constant, its derivative is 0: $$\frac{d}{dx}(1) = 0$$

Step 6: Combine the results

Now, we add or subtract the derivatives of all terms following the sum/difference rule: $$\frac{d}{dx}\left(f(x)\right) = \frac{20}{3}x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2x - 3 + 0$$ So the derivative of the function \(f(x)\) is: $$f'(x) = \frac{20}{3}x^{\frac{1}{3}} - x^{\frac{1}{2}} + 2x - 3$$

Key Concepts

Power Rule

The power rule is a fundamental tool in calculus. It allows us to differentiate functions of the form \(x^n\). The rule states that if \(n\) is any real number, the derivative of \(x^n\) is \(nx^{n-1}\).
This makes the process of finding derivatives straightforward when dealing with polynomial terms.
In our example, when differentiating terms like \(5x^{4/3}\) or \(x^2\), the power rule lets us determine their derivatives quickly.
By multiplying the power by the coefficient and reducing the power by one, we find the new derivative expression.
This simplification is one of the reasons the power rule is a staple in calculus problem-solving.

Derivative of Polynomials

Polynomials are expressions made up of variables and coefficients, involving terms of varying powers.
To find the derivative of a polynomial, we simply differentiate each term individually using the power rule.
In our function, \(f(x) = 5 x^{4 / 3}-\frac{2}{3} x^{3 / 2}+x^{2}-3 x+1\), each term from \(5x^{4/3}\) to the constant \(1\) can be tackled separately.

• Terms involving \(x\) are differentiated using the power rule.
• Constant terms have a derivative of 0, as constants do not change with \(x\).

By applying these rules, we achieve the derivative of the entire polynomial with ease, ensuring we follow each term's individual characteristics.

Sum/Difference Rule

The sum/difference rule is another key concept in differentiation that simplifies the process of dealing with multiple terms.
This rule states that the derivative of a sum or difference of functions is the sum or difference of their derivatives.
This allows us to treat each term of a polynomial separately when finding the overall derivative.
In our example, the derivatives of \(5x^{4/3}\), \(-\frac{2}{3}x^{3/2}\), \(x^2\), \(-3x\), and the constant term \(1\) are individually calculated.
Then, we compile these derivatives together while respecting the signs (addition or subtraction) initially present in the polynomial.
This enables us to piece the derivatives together correctly to form the final expression.

Calculus Problem Solving

Solving calculus problems, especially involving derivatives, requires understanding both the rules and the specific forms of functions given.
By breaking down the problem into manageable steps, such as applying the power rule and sum/difference rule, students can approach problems systematically.
In our function, each step methodically applies learned rules to extract the derivative information effectively.
Employing such structured methods not only aids in solving the problem at hand but also strengthens one's ability to apply calculus to diverse functions and equations.
Each layer of this process, from dissecting the problem to combining separate derivatives, enhances problem-solving skills and helps build confidence in tackling more complex calculus challenges.